Answer:
[tex]\sqrt[3]{27}[/tex]
Step-by-step explanation:
Let's split the cube root of 27 to make this easier:
[tex]\sqrt[3]{27}=\sqrt[3]{3\cdot9}=\sqrt[3]{3\cdot3\cdot3}\\\text{Now, let's put }3\cdot3\cdot3 \text{ in exponential form. We would have:}\\\sqrt[3]{3^3}\\\text{Now, the 3 and 3 and the square root symbol would cancel out, and we }\\\text{ would be left with 3. And since three is an integer, we know that} } \sqrt{27} \text{ is perfect }[/tex]
